Question

A large wooden turntable in the shape of a flat disk has a radius of 2.50 m and a total mass of 130 kg. The turntable is initially rotating at 3.30 rad/s about a vertical axis through its center. Suddenly, a 74.0 kg parachutist makes a soft landing on the turntable at a point on its outer edge.

A) Find the angular speed of the turntable after the parachutist lands. (Assume that you can treat the parachutist as a particle.)B) Compute the kinetic energy of the system before the parachutist lands.C) Compute the kinetic energy of the system after the parachutist lands.D) Why are these kinetic energies not equal?

Answer 1

a) 1.54 rad/s

b) 2212 J

c) 1030J

Step-by-step explanation:

Given:

Radius 'r' = 2.5 m

Total mass of turnable 'M'= 130 kg

Speed of rotation of turntable ' w' = 3.3 rad/s

Mass of parachutist 'm' = 74 kg

A) Assuming the system of turntable and the parachutist. As we know that the system of turn table and parachutist is isolated and there is no external force on this system.

By applying law of conservation of momentum i.e the angular momentum is the same before and after landing. Therefore,

`I_(1)`
`w_(1)`=
`I_{2`
`w_{2`

Where,

`I_(1)` is moment of inertia of the disk and
`I_{2` is moment of inertia of the parachute and the turntable

`I_(1)`= 1/2MR² => 1/2 x 130 x 2.5²

`I_(1)`= 65 x 6.25

`I_(1)`= 406.25 kgm²

`I_{2` = 1/2MR² + mR² => 406.25 + (74 x 2.5²)

`I_{2` = 406.25 + 74 x 6.25 =>406.25 + 462.5

`I_{2` = 868.75 kgm²

`I_(1)`
`w_(1)`=
`I_{2`
`w_{2`

`w_{2` =
`I_(1)`
`w_(1)`/
`I_{2`

`w_{2` = (406.25 x 3.3) / 868.75 =>1340.625 / 868.75

`w_{2` = 1.54 rad/s

B) For initial kinetic energy :

`k_{1` = 1/2
`I_{1`
`w_{1`² => 1/2 x 406.25 x 3.3²

`k_{1` = 1/2 x 406.25 x 10.89

`k_{1` = 2212 J

C)For final kinetic energy:

`k_{2` = 1/2
`I_{2`
`w_{2`² => 1/2 x 868.75 x 1.54²

`k_{2` = 1/2 x 868.75 x 2.3716

`k_{2` = 1030 J

D) Because of the negative work done by the parachutist on the turntable, during the softlanding we have different KE. Energy is decreased due to friction between parachutist and the disc.

Answer 2

a) 1.54 rad/s

b) 2212 J

c) 1030J

Step-by-step explanation:

Given that

Radius of turntable, r = 2.5 m

Total mass of turntable, M = 130 kg

Speed of rotation of turntable, w = 3.3 rad/s

Mass of parachutist, m = 74 kg

To solve this, first, we make an assumption:

That the turntable and the parachutist are a system. Then we use law of conservation of momentum i.e the angular momentum is the same before and after landing. Thus,

I1w1 = I2w2

I1 = moment of inertia of the disk

I1 = 1/2MR²

I1 = 1/2 * 130 * 2.5²

I1 = 65 * 6.25

I1 = 406.25 kgm²

I2 = moment of inertia of the parachute and the turntable

I2 = 1/2MR² + mR²

I2 = 406.25 + 74 * 2.5²

I2 = 406.25 + 74 * 6.25

I2 = 406.25 + 462.5

I2 = 868.75 kgm²

I1w1 = I2w2

w2 = I1w1 / I2

w2 = (406.25 * 3.3) / 868.75

w2 = 1340.625 / 868.75

w2 = 1.54 rad/s

k1 = 1/2.I1.w1²

k1 = 1/2 * 406.25 * 3.3²

k1 = 1/2 * 406.25 * 10.89

k1 = 2212 J

k2 = 1/2.I2.w2²

k2 = 1/2 * 868.75 * 1.54²

k2 = 1/2 * 868.75 * 2.3716

k2 = 1030 J

The KE is different because of the negative work done by the parachutist on the turntable, during the softlanding